On ordinals accessible by infinitary languages
Tom 186 / 2005
Streszczenie
Let $\lambda$ be an infinite cardinal number. The ordinal number $\delta(\lambda)$ is the least ordinal $\gamma$ such that if $\phi$ is any sentence of $L_{\lambda^+\omega}$, with a unary predicate $D$ and a binary predicate $\prec$, and $\phi$ has a model ${\cal M}$ with $\langle D^{\cal M},\prec^{\cal M}\rangle$ a well-ordering of type $\ge\gamma$, then $\phi$ has a model ${\cal M}'$ where $\langle D^{{\cal M}'},\prec^{{\cal M}'}\rangle$ is non-well-ordered. One of the interesting properties of this number is that the Hanf number of $L_{\lambda^+\omega}$ is exactly $\beth_{\delta(\lambda)}$. It was proved in \cite{BarwiseKunen1971} that if $\aleph_ 0 < \lambda < \kappa$ are regular cardinal numbers, then there is a forcing extension, preserving cofinalities, such that in the extension $2^ \lambda = \kappa$ and $\delta (\lambda) < \lambda^{{+}{+}}$. We improve this result by proving the following: Suppose $\aleph_ 0 < \lambda < \theta \leq \kappa$ are cardinal numbers such that
$\bullet$ $\lambda^{< \lambda} = \lambda$;
$\bullet$ ${\rm cf}( \theta) \geq \lambda^+$ and $\mu^\lambda < \theta$ whenever $\mu < \theta$;$\bullet$ $\kappa^\lambda = \kappa$.
Then there is a forcing extension preserving all cofinalities, adding no new sets of cardinality $< \lambda$, and such that in the extension $2^ \lambda = \kappa$ and $\delta( \lambda) = \theta$.
